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Math 122 - Calculus for Biology II |
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San Diego State University -- This page last updated 21-Oct-00 |
Introduction to Differential Equations - Examples
This section introduces a few problems in differential equations. Mostly, you will use you knowledge about differentiation to show that a given solution satisfies the differential equation. Later in the course, we will learn the techniques to solve the differential equations.
Example 1: Suppose that a culture of Escherichia coli is growing according to the Malthusian growth model
It is found that the population doubles in 25 minutes. Find the growth rate constant and the solution to this differential equation. Finally, compute the population after one hour.
Solution: From the lecture notes, we see that the general solution to this problem is given by
If the population doubles in 25 minutes, then
P(25) =200000 = 100,000
e25r.
Taking the logarithm of both sides, ln(2)
=25r, so the growth rate constant
The specific solution is given by
P(t) = 100,000
e0.0277t,
where t is in minutes.
one hour is 60 minutes, so the population after
one hour is P(60) = 100,000
e0.0277(60) = 527,803
Example 2: a. A mass suspended by a spring without resistence satisfies the second order linear differential equation
Show that two of the solutions to this differential equation are given by
b. If this mass experiences experiences certain damping forces, then satisfies the differential equation
Show that
is a solution to this differential equation. Graph the solution to this differential equation and describe the behavior of the mass.
Solution: a. One method of solving this problem is using the solution given in the lecture notes where k2 = 5, so the frequency of the sine and cosine functions, k = . The direct manner to solve this problem is to take two derivatives of the solutions and substitute them into the differential equation. Thus,
Substituting these solutions into the differential equation, it follows that
These show that the given expressions are solutions to the differential equation for free motion of a spring-mass system.
b. To verify that y(t) = 2e-tsin(2t) is a solution to the given differential equation for a damped oscillator, we again take two derivatives of the expression. This gives:
y''(t) = 2e-t(-4sin(2t)-2cos(2t)) - 2e-t (2cos(2t)-sin(2t)) = -2e-t (4cos(2t)+3sin(2t)).
We substitute these expressions into the differential equation and obtain:
By combining like terms, everything cancels, so y"(t) + 2y'(t) + 5 y(t) = 0.
A graph of the displacement of the mass over two periods of the trigonometric functions is given by
From the graph you can see that the mass is moving up and down, but the amplitude of the oscillation gets smaller exponentially with passing time, rapidly converging to zero.
Visually, we saw on the lecture page that the solution of the free motion and the damped motion are represented by the following animated .gif of Professor Dan Russell:
Example 3: Animals lose moisture proportional to their surface area. If V(t) is the volume of water in the animal, then the moisture loss satisfies the differential equation
where the initial condition V(0) = 8 cm3 is the initial amount of water in the animal and t is measured in days. Show that the solution to this differential equation, including the initial condition, is given by
Sketch a graph of the solution and determine when the animal becomes totally dessicated according to this model.
Solution: By substitution V(0) = (2 - 0.01(0))3 = (2)3 =8, so the initial condition is satisfied. To show that V(t) is a solution of the differential equation, we differentiate this expression. Thus,
But V 2/3(t) = (2 - 0.01t)2, so when we substitute this into the expression for the derivative, we obtain
which is the differential equation. A graph of the solution is below.
To find the time when total dessication occurs, we must solve
Thus, 2 - 0.01t = 0 or t = 200 days.
Example 4: Consider the nonautonomous differential equation with initial condition (initial value problem):
Show that the solution to this differential equation, including the initial condition, is given by
Sketch a graph of the solution.
Solution: Again we begin by evaluating y(t) at t = 0. By substitution, y(0) = 2(02+1)-1 = 2(1)-1 = 2, so the initial condition is satisfied.
Next we differentiate y(t). We write y(t) = 2(t2+1)-1, then apply the chain rule to this expression. The result is
However,
so the differential equation is satisfied. Below is a graph of the solution.